$\lim_{n\rightarrow\infty}n(\ln n)a_n=0$ implies $\sum a_n$ converges?

Question

Is it true that If $$\lim_{n\rightarrow\infty}n(\ln n)a_n=0,$$ then the series $$\sum_{n=1}^\infty a_n$$ converges?

If so, I want to know the proof. If not, I want to know the counter example.

Answer 1

The series $$\sum_{n=4}^\infty \frac{1}{n\log n \log\log n}$$ does not converge. Use the integral test.

Answer 2

Assume that $\sum_{n=1}^\infty b_n$ is a divergent positive series. Then you can always find another divergent positive series $\sum_{n=1}^\infty a_n$ with the property that $$ \lim_{n\to\infty} \frac{a_n}{b_n} = 0. $$ (In your case, $b_n = \dfrac{1}{n\log n}$.)

One way to see this is to use a theorem of Abel and Dini:

Theorem Assume that $\sum_{n=1}^\infty b_n$ is a divergent positive series, and let $B_n = b_1 + \cdots + b_n$ denote its partial sums. Then the series $$ \sum_{n=1}^\infty \frac{b_n}{B_n} $$ also diverges.

Proof First note that $$ \frac{b_{n+1}}{B_{n+1}} + \frac{b_{n+2}}{B_{n+2}} + + \cdots + \frac{b_{n+k}}{B_{n+k}} \ge \frac{b_{n+1}+b_{n+2}+\cdots+b_{n+k}}{B_{n+k}} = 1 - \frac{B_n}{B_{n+k}}. $$ Since we assume that $B_n \to \infty$, for each $n$, we can choose $k_n$ such that $$ \frac{B_n}{B_{n+k_n}} < \frac12, $$ i.e. $$ \frac{b_{n+1}}{B_{n+1}} + \frac{b_{n+2}}{B_{n+2}} + + \cdots + \frac{b_{n+k_n}}{B_{n+k_n}} > \frac12. $$ Summing blocks like these, we see that $$ \sum_{n=1}^\infty \frac{b_n}{B_n} $$ diverges.